Difference between revisions of "1950 AHSME Problems/Problem 20"
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Note that <math>x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)</math>, so <math>x^{13} - 1</math> is divisible by <math>x-1</math>, meaning <math>(x^{13} - 1) + 2</math> leaves a remainder of <math>\boxed{\mathrm{(D)}\ 2.}</math> | Note that <math>x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)</math>, so <math>x^{13} - 1</math> is divisible by <math>x-1</math>, meaning <math>(x^{13} - 1) + 2</math> leaves a remainder of <math>\boxed{\mathrm{(D)}\ 2.}</math> | ||
| + | |||
| + | ===Video Solution=== | ||
| + | https://www.youtube.com/watch?v=z4-bFo2D3TU&list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&index=3&t=3s | ||
| + | (Solution at 41:25 - AMBRIGGS | ||
==See Also== | ==See Also== | ||
Revision as of 10:15, 30 July 2022
Contents
Problem
When 
is divided by 
, the remainder is:
Solution
Solution 1
Use synthetic division, and get that the remainder is 
Solution 2
By the remainder theorem, the remainder is equal to the expression when 
This gives the answer of 
Solution 3
Note that 
, so 
is divisible by , meaning 
leaves a remainder of
Video Solution
https://www.youtube.com/watch?v=z4-bFo2D3TU&list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&index=3&t=3s (Solution at 41:25 - AMBRIGGS
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
